Superposition principle: Worked examples — Physics deep dive

Before anything, let us make sure the symbols are earned.

Definition Amplitude

$A_{res}$ , angular frequency $ω$ , phase difference $ϕ$ , and resultant phase $α$ — in plain words

Amplitude $A$ = the biggest push a single wave can give a point, measured from the flat resting line. Picture the height of a swing at its highest — that height is $A$ .
Angular frequency $ω$ = how fast the wave cycles, measured in radians per second. In $y = A sin (ω t)$ , the quantity $ω t$ is the angle (in radians) the wave has swept since $t = 0$ . One full cycle happens when $ω t$ grows by $2 π$ . It is "angular" because we track the wave as an angle going around a circle (see the phasor idea below).
Phase difference $ϕ$ = how far one wave is shifted along in its cycle compared to the other, measured as an angle. A full cycle is $36 0^{\circ} = 2 π rad$ . If wave 2 starts its up-swing when wave 1 is already a quarter of the way through, wave 2 leads by $ϕ = 9 0^{\circ}$ .
Resultant amplitude $A_{res}$ = the biggest push of the single combined wave you get after superposing the two — the height of the summed wave, not of either original.
Resultant phase $α$ = the direction (phase angle) of the combined wave, measured from wave 1. When you add the two arrows and get one green arrow (Figure s01), $α$ is how tilted that green arrow is away from wave 1. A positive $α$ means the combined wave leads (starts its up-swing earlier than) wave 1; negative means it lags.

We will use one master formula, taken from the parent note and derived by cosine rule on those two arrows:

Intuition Where

$tan α = \frac{A _{2} sin ϕ}{A _{1} + A _{2} cos ϕ}$ comes from — build it yourself Look at Figure s01 and drop the two arrows onto x–y axes.

Wave 1 lies flat along the horizontal, so its components are $(A_{1}, 0)$ .
Wave 2 is tilted up by $ϕ$ , so by "adjacent = length· $cos$ , opposite = length· $sin$ " its components are $(A_{2} cos ϕ, A_{2} sin ϕ)$ .
Adding arrows = adding components. The green resultant therefore has:
- horizontal part $= A_{1} + A_{2} cos ϕ$ (both arrows' rightward pushes),
- vertical part $= A_{2} sin ϕ$ (only wave 2 has a vertical push).
The green arrow makes angle $α$ with the horizontal, and for any arrow $tan (tilt) = \frac{vertical part}{horizontal part}$ (opposite over adjacent). That is exactly the formula: $tan α = \frac{vertical}{horizontal} = \frac{A _{2} s i n ϕ}{A _{1} + A _{2} c o s ϕ} .$ So $tan α$ answers a different question from $A_{res}$ : not "how long is the green arrow?" but "which way does it point?"

The scenario matrix

Every superposition problem you will meet is one of these cells. The examples below are labelled with the cell they hit.

#	Case class	What's special	Example
A	1-D signed sum, same sign	pulses reinforce, no angles	Ex 1
B	1-D signed sum, opposite sign	partial / total cancellation	Ex 1, Ex 2
C	Degenerate: equal & opposite	$A_{res} = 0$ , boundary case	Ex 2
D	In phase, $ϕ = 0$	limiting max, $A_{res} = A_{1} + A_{2}$	Ex 3
E	Out of phase, $ϕ = 18 0^{\circ}$	limiting min, $A_{\text{res}}=	A_1-A_2
F	Quarter phase, $ϕ = 9 0^{\circ}$	phasors perpendicular ⇒ Pythagoras	Ex 4
G	General angle $0 < ϕ < 18 0^{\circ}$	full cosine rule needed	Ex 5
H	Angle beyond $18 0^{\circ}$ ( $ϕ > π$ )	sign of $cos$ flips, cos-half trap	Ex 6
I	Resultant phase (direction, not size)	uses $tan α$ / atan2	Ex 7
J	Real-world word problem	translate words → $ϕ$	Ex 8
K	Exam twist: find $ϕ$ from given $A_{res}$	invert the formula	Ex 9
L	Negative phase $ϕ < 0$	lag; same amplitude, flipped $α$	Ex 10

Worked Examples

Cells A, B — 1-D signed sum

Worked example Example 1 — Two pulses meet (same-sign then opposite-sign)

At point P, at one instant pulse 1 gives $+ 6 cm$ and pulse 2 gives $+ 2 cm$ . A moment later pulse 1 gives $+ 6 cm$ and pulse 2 gives $- 2 cm$ . Find the net displacement at each instant.

Forecast: Guess both before reading — do the numbers just add?

Instant 1: $y = y_{1} + y_{2} = 6 + 2 = + 8 cm$ . Why this step? In one dimension a wave's displacement is a signed number (up = $+$ , down = $-$ ). Superposition says the medium does both pushes at once, so we add the signed numbers — no angles needed because there is only one line to move along.
Instant 2: $y = 6 + (- 2) = + 4 cm$ . Why this step? Same rule; the negative sign of pulse 2 means it pushes down while pulse 1 pushes up, so they partially cancel.

Verify: Both answers are smaller-or-equal to the total available push $6 + 2 = 8$ . Same-sign gave the max ( $8$ ), opposite-sign gave less ( $4$ ). Units stay cm throughout. ✓

Cell C — the degenerate cancellation

Cells D, E — the two limiting phases

Worked example Example 3 — In phase vs out of phase (limits of the master formula)

Two waves, $A_{1} = 4$ , $A_{2} = 3$ (same units). Find $A_{res}$ for (a) $ϕ = 0$ , (b) $ϕ = 18 0^{\circ}$ .

Forecast: What is the biggest and smallest a $4$ and a $3$ can make together?

(a) $ϕ = 0$ : convert: $0^{\circ} = 0 rad$ , so $cos 0 = + 1$ , giving $A_{res} = 4^{2} + 3^{2} + 2 (4) (3) (1) = 16 + 9 + 24 = 49 = 7.$ Why this step? At $ϕ = 0$ the two arrows point the same way (in Figure s01 imagine collapsing the red angle to zero so blue and yellow line up): the answer is simply $A_{1} + A_{2} = 7$ . This is the maximum — constructive interference.
(b) $ϕ = 18 0^{\circ}$ : convert: $18 0^{\circ} = π rad$ , so $cos π = - 1$ , giving $A_{res} = 16 + 9 + 2 (4) (3) (- 1) = 25 - 24 = 1 = 1.$ Why this step? At $ϕ = 18 0^{\circ}$ the arrows point opposite ways; the answer is $∣ A_{1} - A_{2} ∣ = ∣4 - 3∣ = 1$ . This is the minimum — destructive interference. It is not zero here because the amplitudes are unequal.

Verify: $A_{res}$ always lands in the band $[∣ A_{1} - A_{2} ∣, A_{1} + A_{2}] = [1, 7]$ . Every other phase gives a value inside this band. ✓

Cell F — quarter phase, perpendicular phasors

Worked example Example 4 — Ninety degrees ⇒ Pythagoras

$y_{1} = 4 sin ω t$ , $y_{2} = 3 sin (ω t + 9 0^{\circ})$ , where $ω$ is the shared angular frequency (both waves cycle at the same rate, so their phasors spin together and the difference $ϕ = 9 0^{\circ}$ stays fixed). Find $A_{res}$ .

Forecast: The arrows are at right angles — what shape does that make?

Draw the phasors: arrow 1 (length 4) along the reference, arrow 2 (length 3) turned $9 0^{\circ}$ from it. Why this step? A $9 0^{\circ}$ phase difference means the second arrow points straight up relative to the first — they are perpendicular. (Because both share $ω$ , spinning the whole picture does not change this right angle.)
Tip-to-tail they form a right triangle with legs 4 and 3. Convert: $9 0^{\circ} = \frac{π}{2} rad$ , and $cos \frac{π}{2} = 0$ , so the cosine term vanishes: $A_{res} = 4^{2} + 3^{2} = 25 = 5.$ Why this step? For a right angle, the law of cosines loses its cosine term ( $cos 9 0^{\circ} = 0$ ) and reduces to plain Pythagoras.

Verify: Check against the master formula: $16 + 9 + 2 (4) (3) cos \frac{π}{2} = 25 + 24 (0) = 5$ . ✓ Falls inside the band $[1, 7]$ . ✓

Cell G — a general in-between angle

Cell H — phase beyond $18 0^{\circ}$ (the cos-half trap)

Worked example Example 6 — When

$ϕ > 18 0^{\circ}$ Two equal waves, $A = 2$ , with phase difference $ϕ = 24 0^{\circ}$ . Find $A_{res}$ .

Forecast: People often plug into $2 A cos (ϕ /2)$ and get a negative number — is that allowed?

Equal amplitudes, so use $A_{res} = 2 A cos (ϕ /2)$ . Convert: $ϕ = 24 0^{\circ} = \frac{4 π}{3} rad$ , so $ϕ /2 = 12 0^{\circ} = \frac{2 π}{3} rad$ , and $cos \frac{2 π}{3} = - 0.5$ : $2 (2) cos (\frac{24 0 ^{\circ}}{2}) = 4 cos 12 0^{\circ} = 4 (- 0.5) = - 2.$ Why this step? The formula gives $- 2$ , but amplitude cannot be negative — a length is never below zero.
Fix — take the magnitude: the physical amplitude is $A_{res} = ∣ - 2∣ = 2$ . Why this step? $cos (ϕ /2)$ went negative because $ϕ /2 = 12 0^{\circ}$ is past $9 0^{\circ}$ , where cosine crosses from $+$ to $-$ . A negative $cos (ϕ /2)$ does not mean a negative length — length is always positive. What the minus sign actually encodes is that the resultant arrow has flipped to point the other way (an extra $18 0^{\circ}$ has been added to the resultant's own phase). So we read off the amplitude as the absolute value $∣2 A cos (ϕ /2) ∣$ , and remember the flip separately if we need the direction.
Safer route — use the sign-stable master formula, which never produces a spurious minus because the whole expression sits under a square root: $A_{res} = 2^{2} + 2^{2} + 2 (2) (2) cos \frac{4 π}{3} = 4 + 4 + 8 (- 0.5) = 8 - 4 = 4 = 2.$ Why this step? Here $cos ϕ$ (not $cos ϕ /2$ ) enters, and the square root forces a non-negative answer automatically — so for $ϕ > 18 0^{\circ}$ prefer this form and you sidestep the trap entirely.

Verify: Both routes agree: magnitude $2$ . ✓ And $2$ sits inside the band $[∣2 - 2∣, 2 + 2] = [0, 4]$ . ✓

Cell I — the resultant's direction (phase), not its size

Worked example Example 7 — Where does the summed arrow point?

$y_{1} = 4 sin ω t$ , $y_{2} = 3 sin (ω t + 9 0^{\circ})$ . Find the resultant phase $α$ (the tilt of the green arrow relative to wave 1).

Forecast: The bigger arrow (4) points along the reference; the smaller (3) tilts it up — expect a modest angle under $4 5^{\circ}$ .

Use $tan α = \frac{A _{2} sin ϕ}{A _{1} + A _{2} cos ϕ}$ — the formula we built from components above. Convert: $ϕ = 9 0^{\circ} = \frac{π}{2} rad$ , so $sin \frac{π}{2} = 1$ and $cos \frac{π}{2} = 0$ : $tan α = \frac{3 s i n 9 0 ^{\circ}}{4 + 3 c o s 9 0 ^{\circ}} = \frac{3 ( 1 )}{4 + 0} = \frac{3}{4} = 0.75.$ Why this step? Numerator = the vertical part of the green arrow (only wave 2 contributes up); denominator = the horizontal part (wave 1 plus wave 2's sideways bit). Their ratio is opposite-over-adjacent = the tilt. This answers "which direction?", a different question from Example 4's "how long?".
Here the horizontal part $A_{1} + A_{2} cos ϕ = 4$ is positive, so plain $arctan$ is safe (no quadrant flip). $α = arctan (0.75) \approx 36.8 7^{\circ}$ . Why this step? $arctan$ asks "which angle has tangent $0.75$ ?" — it undoes the tan. Both parts positive ⇒ first quadrant ⇒ atan2(3,4) agrees with arctan(0.75), no correction needed.

Verify: The full wave is $y = 5 sin (ω t + 36.8 7^{\circ})$ . Check the size matches Ex 4 ( $5$ ) and the angle is under $4 5^{\circ}$ as forecast (because wave 1 is the longer arrow, it "wins" the direction). ✓

Cell J — real-world word problem

Worked example Example 8 — Two loudspeakers (translate words →

$ϕ$ ) Two loudspeakers play the same pure tone in phase, each producing sound of amplitude $A = 1$ (arbitrary units) at your ear. Because you stand closer to one, its sound travels an extra half-wavelength ( $λ /2$ ) before reaching you. What amplitude do you hear?

Forecast: Half a wavelength extra — louder or dead-silent?

Convert path difference to phase: a full wavelength $λ$ is one full cycle $= 36 0^{\circ} = 2 π rad$ . So an extra $λ /2$ is $ϕ = \frac{λ /2}{λ} \times 2 π rad = π rad = 18 0^{\circ} .$ Why this step? Phase difference $ϕ = \frac{2 π}{λ} \times (path difference)$ — extra distance travelled is time travelled is phase accumulated. This is the bridge from geometry (metres) to phase (radians).
Equal amplitudes, $ϕ = π rad$ : $A_{res} = 2 A cos (π /2) = 2 (1) (0) = 0.$ Why this step? $cos (π /2) = cos 9 0^{\circ} = 0$ , so total cancellation — a "dead spot".

Verify: Master formula: $1 + 1 + 2 (1) (1) cos π = 2 - 2 = 0$ . ✓ You hear silence there — a real, hearable effect in Interference of waves. Energy isn't lost; it is louder elsewhere.

Cell K — exam twist: invert the formula

Worked example Example 9 — Given the answer, find the phase

Two equal waves of amplitude $A = 3$ combine to give $A_{res} = 3$ . What is the phase difference $ϕ$ (smallest positive value)?

Forecast: The result equals a single wave's amplitude — the two must be partly fighting. Guess whether $ϕ$ is under or over $9 0^{\circ}$ .

Set the equal-amplitude formula equal to the given result: $2 A cos (\frac{ϕ}{2}) = A_{res} \Rightarrow 2 (3) cos (\frac{ϕ}{2}) = 3 \Rightarrow cos (\frac{ϕ}{2}) = 0.5.$ Why this step? We know the output and one input; we solve backwards for $ϕ$ . This is why $arccos$ exists — it asks "which angle has this cosine?"
$\frac{ϕ}{2} = arccos (0.5) = 6 0^{\circ} = \frac{π}{3} rad \Rightarrow ϕ = 12 0^{\circ} = \frac{2 π}{3} rad .$ Why this step? Multiply back by 2. Since $cos$ is positive and we want the smallest positive angle, $6 0^{\circ}$ is the principal value.

Verify: Plug back: $2 (3) cos 6 0^{\circ} = 6 (0.5) = 3 = A_{res}$ . ✓ And $12 0^{\circ} > 9 0^{\circ}$ , matching the forecast that the waves partly oppose.

Cell L — negative phase difference

Worked example Example 10 — Wave 2 lags (

$ϕ < 0$ ) $y_{1} = 4 sin ω t$ , $y_{2} = 3 sin (ω t - 9 0^{\circ})$ , so $ϕ = - 9 0^{\circ}$ (wave 2 lags wave 1). Find $A_{res}$ and the resultant phase $α$ .

Forecast: Compared to Example 7 (lead of $+ 9 0^{\circ}$ ), will the loudness change? Will the direction change?

Amplitude: convert: $- 9 0^{\circ} = - \frac{π}{2} rad$ , and since $cos$ is even, $cos (- \frac{π}{2}) = cos \frac{π}{2} = 0$ , so $A_{res} = 4^{2} + 3^{2} + 2 (4) (3) (0) = 5.$ Why this step? Flipping the sign of $ϕ$ does not change $cos ϕ$ , so the amplitude is identical to Example 7 — a lag sounds exactly as loud as a lead.
Direction: convert: $sin (- \frac{π}{2}) = - 1$ , $cos (- \frac{π}{2}) = 0$ , so $tan α = \frac{3 s i n ( - \frac{π}{2} )}{4 + 3 c o s ( - \frac{π}{2} )} = \frac{3 ( - 1 )}{4 + 0} = - 0.75.$ The horizontal part $4$ is positive, so atan2(-3, 4) gives $α = - 36.8 7^{\circ}$ . Why this step? $sin$ is odd ( $sin (- \frac{π}{2}) = - sin \frac{π}{2}$ ), so the vertical part flips sign while the horizontal part stays — the resultant tilts down instead of up. Same size, mirrored direction.

Verify: $A_{res} = 5$ matches Example 7. The phase $α = - 36.8 7^{\circ}$ is exactly the negative of Example 7's $+ 36.8 7^{\circ}$ — the mirror image, as expected for $ϕ \to - ϕ$ . ✓

Forecast-then-Verify

Recall Forecast: waves of amplitude 5 and 5 with

$ϕ = 12 0^{\circ}$ . Predict $A_{res}$ , then check. Forecast: partly opposing, so somewhere between 0 and 10, below 5. Verify: convert $12 0^{\circ} = \frac{2 π}{3}$ , so $2 (5) cos 6 0^{\circ} = 10 (0.5) = 5$ . Exactly 5 — the crossover point. ✓

Active Recall

Recall Which cell needs no angles at all?

Cells A, B, C — pure 1-D signed addition of displacements.

Path difference of

λ /2

corresponds to what phase difference?

18 0^{\circ} = π rad

(half a cycle).

Why can

2 A cos (ϕ /2)

come out negative, and what do you do?

Because

cos

goes negative past

9 0^{\circ}

; the minus flags a

18 0^{\circ}

flip in direction, so take the absolute value (or use the sign-safe

form) for the amplitude.

For

ϕ = 9 0^{\circ}

the amplitude formula reduces to what?

Pythagoras,

A_{res} = A_{1}^{2} + A_{2}^{2}

, because

cos 9 0^{\circ} = 0

What band must

A_{res}

always lie in?

Between

∣ A_{1} - A_{2} ∣

and

A_{1} + A_{2}

Which formula gives the direction of the resultant, and where does it come from?

tan α = \frac{A _{2} sin ϕ}{A _{1} + A _{2} cos ϕ}

, from vertical-over-horizontal components of the summed arrow; use atan2 when the denominator is negative.

Does flipping the sign of

ϕ

change the loudness?

No —

cos

is even, so

A_{res}

is unchanged; only

α

flips sign (because

sin

is odd).

What is

ω

y = A sin (ω t)

The angular frequency (rad/s);

ω t

is the angle swept, one cycle per

2 π

What does the resultant phase

α

mean physically?

The tilt/direction of the combined wave relative to wave 1 — positive means it leads, negative means it lags.

Connections

Interference of waves — Ex 8's dead-spot is interference in the flesh.
Phasor method — every cell here is one arrow-sum picture.
Beats — what happens when $ϕ$ drifts in time (frequencies differ).
Standing waves — the $ϕ = 18 0^{\circ}$ cancellation, fixed in space.
Simple Harmonic Motion — each wave at a point is an SHM; we added SHMs.
Superposition principle — the parent rule these examples exercise.